Merge Sort Algorithm

Hello people…! In this post I will talk about another sorting algorithm, the Merge Sort Algorithm, which is considered to be the fastest sorting technique by many. Merge Sort was invented by a famous computer scientist John von Neumann. It is a stable sort, so the relative position of the elements are preserved. It uses O(n) auxiliary memory to sort the elements in O(N log N) time. It is the classic example of Divide-and-Conquer Algorithms. It is very easy to understand and code. So, let’s get started..!

In the Merge Sort, we divide the array into halves recursively until we cannot divide further. Then, we merge them into a single array which would be sorted. The partitioning is similar to that in Quick Sort. We can put the merge part of the algorithm in a step-by-step process –

  • Create an array which is large enough to accommodate all the elements of the two arrays to be merged.
  • Compare the first elements of both arrays. Pick the least one and put it into the array we just created.
  • Now we compare the first element of one array and the second element of the other array (from which the element was picked). We choose the smaller element and put it into the large array.
  • We do this until we run out of elements. Then, our larger array is sorted.

For a better understanding of the merge procedure, look at the sketch below –

Merge Sort Step-by-Step

Merge Sort Step-by-Step

This is the way we merge two arrays. As you can see the prerequisite for this procedure to work is that the two smaller arrays that we are given must be sorted. How do we ensure this in our algorithm. Remember, at the lowest level, the partitions stop when the array has only one element, which is considered to be sorted. Then, we would merge two arrays of size 1 into a sorted array of size 2. This helps in merging two arrays of size 2 to form an array of size 4. In this way the recursion goes. So we will have the smaller arrays sorted initially.

As we have gained the knowledge about the working of the algorithm. Let’s put it in terms of pseudo-code. It comprises of two functions –

mergeSort(Array, low, high)
	if low < high
		mid = (low + high) / 2

		for i = low to mid
			LeftPartition[i - low] = Array[i]

		for i = mid + 1 to high
			RightPartition[i - mid - 1] = Array[i]

		mergeSort(LeftPartition, low, mid)
		mergeSort(RightPartition, mid + 1, high)
		mergePartitions(Array, low, high, LeftPartition, RightPartition)

mergePartitions(Array, low, high, LeftPartition, RightPartition)
	i = j = k = 0
	mid = (low + high) / 2

	while (i <= low - mid && j < high - mid)
		if (LeftPartition[i] < RightPartition[j])
			Array[k] = LeftPartition[i]
			Array[k] = RightPartition[j]

	// If any array is left out
	if i > low - mid && j < high - mid
		while j < high - mid
			Array[k] = RightPartition[j]
		while i <= low - mid
			Array[k] = RightPartition[i]

The pseudo-code is clear and self-explanatory. If you have any doubts, feel free to comment them. Looking at the pseudo-code try to code Merge Sort algorithm. If you get any doubts, keep reading the explanation again and again. I’m sure you’ll get it if you give your best. If you didn’t, don’t fret, I posted my code below –


I hope you get the intuition behind the Merge Sort Algorithm. It is the best example to explain Divide-and-Conquer technique. It has a runtime complexity of O(N log N). Which can be proved easily by using a Recursion Tree –

Recursion Tree for Merge Sort Algorithm

Recursion Tree for Merge Sort Algorithm

Optimisation – Merge Sort can be optimised to perform better. The idea behind this is the asymptotic complexities of various sorting algorithms. When we say that Merge Sort is better than Insertion Sort because it has a better worst case of O(n log n) which is less than O(n2), this is difference can be seen only for large values of ‘n’. This is because, the terms like O(n log n), O(n) and all are valid only when n → ∞ . So, for small values of ‘n’, like 100, we may not find any significant performance differences between Merge Sort and Insertion Sort. For smaller ‘n’ an optimised Bubble Sort can actually out perform the Merge Sort, because Merge Sort comes with a lot of overheads, the partition, the merging, the recursive calls and so. Whereas, the Bubble Sort has no such over heads. So, how about we apply an optimised Bubble Sort for smaller sized partitions inside the algorithm of Merge Sort…? That is how we optimise the Merge Sort.

But how do we decide how long this “smaller sized partitions” should be..? One way is you could do an analysis. You can just test your optimised Merge Sort for various lengths of cut-off’s. You need to run your code for the same input and keep varying the ‘threshold’ value below which the smaller sized partitions are sorted by optimised Bubble Sort and not by Merge Sort partitions. You could actually plot a graph between threshold and runtime, where you will find a minimum runtime for a certain value of threshold. But remember, if you want to see significant results you need to input as many as 105 or 106 values.

The next best thing you can do is make a guess by looking at your resources. The idea is to look at your cache. Let’s say your cache is 2MB. Then, your cache can accommodate an integer array of size 5 ✗ 105. But, the whole of cache will not be available to you, so let’s say when you operate on an array of size 100, it is most likely that it’ll end up being in the cache. So, if we do an optimised Bubble Sort for arrays of size 100, we are most likely doing that in cache and as our process has less overheads, it actually increases the performance.

Its okay if your don’t get the whole threshold concept. But the gist of the discussion is that we will use an optimised Bubble Sort for smaller sized partitions in Merge Sort. Try to code this weird practice. You only have to make minor adjustments to your code. I’m sure you’ll get it if you give a little serious thought. I will post the code very shortly, until then keep trying!

This is the Merge Sort Algorithm. If you have any doubts or anything to say, feel free to comment them. I hope my post helped you in getting an idea of the Merge Sort. If it did, let me know by commenting…! Keep practising… Happy Coding…! 🙂

Segment Trees

Hello people…! In this post I will talk about one of the most versatile data structure in competitive programming, the Segment Tree. Almost in every contest that you do in the major coding websites, at least one question is asked which uses a Segment Tree in its solution. This is because it is easy to code, and does some useful operations quickly, in logarithmic time. Segment Trees are used where the problem demands these two queries on a given array arr[1 … n] –

  1. Sum of elements in the range arr[L … R]. Where, 1 <= L <= R <= n.
  2. Updating an element arr[i].

You can think of doing this in the naive approach where you directly operate on the input array itself. You could do the first operation in O(n) time by running a loop calculate the sum. The second operation could be done in O(1) time, because, it’s only simply updating. This is good if the number of sum queries asked are very less and the update queries are more.

The next best thing you could think of is, Prefix Sum Array. The Prefix Sum Array is an array where the element arr[i] holds the sum of values from arr[1] to arr[i]. You can construct the Prefix Sum Array in O(n) time and solve the first query in O(1) time. Prefix Sum Array is one of my favourite techniques to find out the sum in a given range. But the update query takes O(n) time. If you want to update the ith element, you’d have to update all the elements from i to n, which takes O(n) in the worst case.

How about doing both of them in O(log n) time…? That’s really quick..! This is the speciality of the Segment Tree. We can get the sum query and the update query done in logarithmic time. The Segment Tree works on the principle of partial sums. If you have read my post on Fenwick Trees (or) Binary Indexed Trees, or if you already know BIT, Segment Tree should be easy for you. For those who don’t know, don’t worry, you’ll have an easier time learning the Binary Indexed Trees than compared to the others. So, let’s get started..!

As I mentioned earlier, the Segment Tree works on the principle of partial sums. What this principle is that we somewhere hold the sum of a few elements and keep building on it. In a Segment Tree, this “partial” is actually “half”. The Segment Tree structurally is a Binary Tree. Every internal node is associated with a range of numbers, whose sum is stored in that node. Now, what we do is partition the array into two halves. The left child will hold the sum of elements in the left partition and the right child will hold the sum of elements in the right partition. You may not get the idea of Segment Tree in the first reading. That’s okay. Give another reading and ponder upon the sketch below for a better understanding.

Structure of Segment Tree

Structure of Segment Tree

I hope it is clear from the picture how the Segment Tree works on the partial sums, which here is the left partition sum and the right partition sum. The leaf nodes represent the individual elements of the input array.

Note that the array we will not store the array in the nodes of our Segment Tree as was depicted in the sketch. This was done in the sketch only to give you the idea. The array will not be stored in the nodes. However, we will store the range limits on which the sum is defined, i.e, we will store the start index and the end index of the span of elements in the input array whose sum is there in the node. We will use a struct for the node which looks like this –

struct node
	int value, low, high;
	struct node * leftChild;
	struct node * rightChild;

The variable ‘value’ was to store the sum, ‘low’ and ‘high’ is to store the range. The variables ‘leftChild’ and the ‘rightChild’ are to point to the children.

Building the Segment Tree –

The Segment Tree is built in a recursive manner. We could write it in a step-by-step procedure –

  • Allocate memory.
  • Start with the root.
  • Partition the array you are given into two halves.
  • Create nodes for each half.
  • The value of the parent node is the sum of values on its children.
  • This is calculated by recursively calling the procedure on the children.
  • When the array given reduces to one element. The procedure must create a node and return the value.
  • When the parent node has acquired the sum of its children, it must return its current value to carry on the recursion.

The step-by-step procedure may not be clear in the first reading. You will know what I was talking about when you understand the pseudo-code below –

int buildSegmentTree(Array, low, high, SegmentTree)
    if low == high
        SegmentTree->value = Array[low]
        SegmentTree->low = SegmentTree->high = low
        return SegmentTree->value
    mid = (low + high) / 2
    SegmentTree->low = low
    SegmentTree->high = high
    SegmentTree->leftChild = allocate memory using calloc()
    SegmentTree->rightChild = allocate memory using calloc()
    SegmentTree->value = buildSegmentTree(Array, low, mid, SegmentTree)
                       + buildSegmentTree(Array, mid + 1, high, SegmentTree)
struct node * getSegmentTree(Array, ArraySize)
    if ArraySize <= 0
        return NULL
    struct node * SegmentTree = allocate memory using calloc()
    buildSegmentTree(Array, 1, ArraySize, SegmentTree)
    return SegmentTree

I made the pseudo-code a little extra-descriptive so that you can understand the step-by-step procedure and the code simultaneously. If you have any doubts, feel free to comment them.

The time complexity of the code is O(n). This is because in the worst case we would end up building a complete binary tree which will have 2 ✗ n – 1 nodes, and we preform calculations that take O(1) time for every node.

Sum Query –

Now, it is time to discuss the two important queries that we started our discussion with. We will begin with the sum query. We are supposed to tell the sum of all the elements in a given range. Now we will use our ‘low’ and ‘high’ variables in each node of the Segment Tree. The way we do this query mimics the Binary Search. We would be given a well defined range and we will keep comparing that with the range in each node of the Tree. If it fits perfectly, then we simply return the value stored in that node. If it doesn’t we have a few cases to handle. To understand the cases, consider the following notations –

  • NodeLeft – This is the left bound of the range whose sum is stored in the node. We actually named this as ‘low’ inn our struct.
  • NodeRight – This is the right bound of the range whose sum is stored in the node. We actually named this as ‘high’ inn our struct.
  • QueryLeft – This is the left bound of the range whose sum we are supposed to calculate.
  • QueryRight – This is the right bound of the range whose sum we are supposed to calculate.
  • Mid – This is the index upon which the partition occurs to assign values to the children nodes. It is essentially the mean of NodeLeft and NodeRight.
  • Value – This is the value that the current node is holding. Same as the ‘value’ variable in our struct.

Now, let’s talk about the cases we would encounter –

  1. (NodeLeft == QueryLeft && NodeRight == QueryRight) – This is the best thing that can happen. As we discussed a moment ago, we simply return the Value.
  2. (NodeLeft >= QueryLeft && NodeRight >= QueryRight) – This is the situation where among the elements that the current node holds the sum of, a few of the elements are not needed. The range that the current node has covers more than that’s needed. Now, to proceed with the recursion, we use Mid –
    • (QueryLeft <= Mid) → getSumQuery(SegmentTree->leftChild, QueryLeft, Mid) – This condition checks if the query demands the sum of elements which are in the left partition. So, we carry on the recursion with query range concerning only the left partition.
    • (QueryRight > Mid + 1) → getSumQuery(SegmentTree->leftChild, QueryLeft, Mid) – This checks if the query wants elements of the right partition. So, we carry on the recursion with query range concerning only the right partition.

    Note that both the conditions must be checked for the sum query to give accurate results. A special case arises when (QueryLeft == QueryRight), which means that we are looking for one element, which would obviously be either in the left or the right partition. In that case, we don’t have the necessity of  checking the second condition if the first condition is already met. This is because if we know the query asks for an element in the left partition, we have nothing to do with the right partition in this case.

This will be more clear once you put this idea to work by pen-and-paper. Coding this is simple too. Once you are able to code it, you will realise its resemblance to Binary Search. In the worst case, we would have to travel all the way down the tree, which takes O(log n).

Update Query –

What do we do if we were to update an element initially given in the input. How would we have to modify our Segment Tree so that it is still correct? We know that the initial input elements can be found at the leaves of the Segment Tree. So, we should be updating that. How do we traverse down to that leaf? We do this similarly as we did it in Sum Query, where we took a range and went down the tree based on it. If we put it in Sum Query terms, QueryLeft is equal to QueryRight. We search for it in a similar fashion. Then, we update the node which we came searching for. We also keep a note of the difference, and we add the difference to all those nodes which we came by. This is because, when you update a leaf of the Segment Tree, the internal nodes which would get affected are those which contain the partial sum in which the leaf is involved. We already met those nodes in our recursive traversal. When we are “coming out” of recursion, we update them by the difference that was caused by the new value. So, basically we go down the tree and come back. So, this operation takes O(2 log n) which is O(log n).

Take a break for a while and come back to start coding the Segment Tree. You will need to do a lot of pen-and-paper work while coding this data structure. And it is totally worth it. To print your Segment Tree, you could use the Tree Walks, In-order or Pre-order or whichever pleases you. If you are getting stuck, you can always refer to my code below –


There are many varieties of questions that can be asked based on this Data Structure. I’m sure you’ll start liking and appreciating this data structure as you keep solving many questions on it. If you have any doubts, feel free to comment them. I hope my post has helped you in understanding the Segment Tree. If it did, let me know by commenting…! Keep practising… Happy Coding…! 🙂

Quick Sort Algorithm

Hello people…! In this post I will talk about the Quick Sort Algorithm. This is an in-place sorting algorithm, which means it works on the given array itself and does not need any additional space, which means less overheads. But this is an unstable sorting algorithm, which means that the relative position of equal elements may not be maintained. So, when an unstable sorting is fine, quick sort is a very good and robust solution. So, lets’s get started with Quick Sort…!

Quick Sort is a Divide-and-Conquer Algorithm. What it means is, the given problem is divided into many sub-problems and we solve the sub-problems to get the solution of the actual problem. Here, we will partition the array into two and work our algorithm on the two arrays and then recursively repeat the same for other partitioned arrays. The process is –

  1. Choose any element in the array. Make it what is called the pivot.
  2. Swap the pivot to the last element.
  3. Make a two way scan of the remaining elements.
  4. Swap elements such that all the elements less than pivot are to the left-end and all the elements greater than pivot are to the right end.
  5. Elements equal to the pivot can go to the either side.
  6. Swap the pivot to the appropriate position.
  7. Apply the same technique to the left sub-array and the right sub-array.
  8. This is done in a recursive manner and the base case is when our array is of size 0 or 1. Arrays of size 0 or 1 are considered to be sorted. So we do nothing in that case.

For a better understanding of the algorithm, look at the sketch below –

Quick Sort Algorithm Step-by-Step

Quick Sort Algorithm Step-by-Step

The technique of quick sort is rather weird but it is straight-forward. Go through the step-by-step process a few more times and try to code the quick sort algorithm. You need to be confident with recursion if you want to get this right. If you have succeeded, well done…! If not, don’t worry, look at my code below and try to figure out what are the small silly that things you have missed…!


Integer Overflow – It would like to point out one thing in the code above. In the line 19, we wrote, “int mid = low + ((high – low) / 2)”. One could have written “mid = (low + high) / 2” also, but there is a reason why we specifically write it in the former way. Suppose, the values of low is INT_MAX – 5 (maximum value of integer) and the value of high is INT_MAX. Upon adding them, we get a value that is greater than INT_MAX, thus a value greater than a 32-Bit integer can hold. So, what happens is that, the bits go from “1111111….” to “0000000….” that is they sort of “reset” themselves. This is similar to the Odometer in a car, when the odometer is maxed out, i.e., it has all 9’s then it resets itself to all 0’s. The same thing happens here too. There is a good depiction of this in Wikipedia. So, when we add such large values we actually get a negative value, to which we would be calculating the mean. And when you give a negative index to an array, it is most likely that your code will crash due to Segmentation Fault. Although this theory makes more sense when the indices of the arrays are 16-Bit integers, every programmer must keep this theory in the back of his mind.

The main concern about this post is not only to give you an idea about Quick Sort Algorithm, but we will discuss the complexity analysis and how to make quick sort faster. Quick Sort algorithm has the following complexities –

  • Worst Case Time Complexity – O(n2)
  • Average Case Time Complexity – O(n log2 n)
  • Best Case Time Complexity – O(n log2 n)
  • Space Complexity – O(n)

The O(n) space complexity of quick sort is due to the O(n) recursion calls that consume the stack. Although the complexities are vivid, the runtime of quick sort depends on a few factors –

  1. Choice of Pivot – The optimal choice of pivot in quick sort has troubled the researchers for quite some time. Generally, the pivot must be selected in such a way that it partitions the array into two-halves, or at the least, partitions the array into nearly equal halves. But we would know nothing about this when we partition. So, we are forced to pick a pivot of our choice. Sometimes, due to the pivot, we might have to do more partitions that we would expect. Consider the case of a sorted array. If we always choose the first element of the array, we would end up doing O(n) partitions. Robert Sedgewick, a computer scientist, suggested to take the median of first, middle and the last element as the pivot. When we implement this, with tail recursion, we can reduce the space complexity to O(log n), which is very less compared to O(n).
  2. Method of Partition – What we have seen till now is a 2-way partition. i.e., we partition the array into two and recursively implement our algorithm. We can do a 3-way partition which can reduce the best case time complexity to O(n). We will discuss this shortly.

Dual Pivot Partitioning – Until now, we did a 2-way partition, or we partitioned the array into 2 parts. Now, we will partition the array into 3 parts and recursively call solve for the partitioned sub-arrays. When we had to divide the array into two, we simply took an element and put all the elements left of it into one array and greater than it into another array. What should we do if we want to partition the array into 3 parts? If you think for a minute or two you can easily figure our the principle. We will simply, pick up 2 elements, say A and B, and partition our array into 3 parts such that –

  • First part has all elements < A.
  • Second part has all elements that is > A, and < B.
  • Third Part has elements > B.

We can put this technique in a step-by-step procedure –

  1. We will choose the starting element, say array[low] and the last element, say array[high] as the pivots. Firstly, check if array[low] < array[high], if not, swap.
  2. Take two variables lt, ht, which store the indices and act as pointers. Set (lt ← low + 1), and (ht ← high – 1).
  3. Start iterating from (low + 1) → (high – 1).
  4. If array[i] is less than array[low], swap array[i] and array[lt] and increment i and lt.
  5. If array[i] is greater than array[high], swap array[i] and array[gt] and decrement gt only.
  6. If array[i] lies between array[high] and array[low], simply continue to the next iteration.
  7. Break out of this procedure, once i crosses gt. This is because when, i > gt, all elements of value array[i] will be greater than array[high], the partitioning guarantees this and they are at their proper place.
  8. Lastly, swap array[low] and array[lt – 1], and swap array[ht + 1] and array[high].
  9. Recursively apply the algorithm on the partitioned sub-arrays.
  10. Your base case if there is only one element in the array, which needs no further sorting.

It is a simple procedure. You just need to be careful about using the variables and the base case must be dealt appropriately. Try to code this technique. I’m sure you’ll get it. If you didn’t I have put my code below –


I hope you get the fundamental idea behind the partition. Now, the time complexity of this algorithm is much like the standard procedure, it is O(n log3n) which is slightly more efficient than the standard one which would be O(n log2n) and we know that for a given value of n, log3n is smaller than log2n. But this difference is not significant for small values of n. Some consider the 3-way partition to add more overheads to the standard procedure. You must choose the best technique that fits your demands.

Random Pivot – There is another optimization technique that is often implemented in quick sort. This is making the choice of the pivot random. Instead of picking the first, or last, or middle element as the pivot we pick a random element in the given range of numbers and partition the array as usual. Using a random number generator, we can pick a random index from a range of indices and partition the array accordingly. You can see it in the code below –


Dijkstra’s 3-way partitioning – Dijkstra’s 3-way partitioning is another approach to partition the array into 3 parts. If the pivot element is ‘P’ then Dijkstra’s 3-way partitioning divides the array into elements which are less than P (left partition), elements which are equal to P (middle partition) and elements which are greater than P (right partition). And since the middle partition is just a collection of equal items, it is considered to be sorted and we recursively call the quick sort on the left and the right partitions. This approach performs exceedingly better than the other implementations of the numbers keep repeating very frequently.


If you have any doubts feel free to comment them. Now, as we have learnt so many varieties of Quick Sort variants. Which one do you think is faster? I did a comparison of the Quick Sort variants for a Weekly Discussion on the blog’s FB page. And this was what I got –

Comparison of Quick Sort Variants

As you can expect, the Dual-Pivot worked pretty faster than the rest due to the O(n log3 n) complexity of Dual Pivot partitioning. You might get a doubt, if dual partitioning gave an O(n log3 n) complexity, will an n-partitioning quick sort give O(n logn n) complexity which is O(n). Can we really achieve a linear time complexity algorithm for Quick Sort? Well, it is a nice question, but remember that, when we were doing a dual pivot partitioning, we had the two pivots in sorted order. Because it was only two elements we didn’t actually do any sorting there. So, if you wanted an n-pivot partitioning in Quick Sort, you can do that, but you’d have to have elements in the sorted order, which is nothing but sorting the elements itself. You see? O(n) complexity sounded awesome a few moments back, but now it sounds like nonsense because we know it wont happen!

I hope my post helped you in learning the Quick Sort algorithm and the principles behind it. If it did, let me know by commenting. Happy Coding…! 🙂

Bellman Ford Algorithm

Hello people…! In this post I will talk about another single source shortest path algorithm, the Bellman Ford Algorithm. Unlike Dijkstra’s Algorithm, which works only for a graph positive edge weights, the Bellman Ford Algorithm will give the shortest path from a given vertex for a graph with negative edge weights also. Due to this, the Bellman Ford Algorithm is more versatile, but, it’s speciality comes at a cost. The runtime complexity of Bellman Ford Algorithm is O(|V||E|), which is substantially more than that of Dijkstra’s Algorithm. Sometimes this algorithm is called Bellman Ford Moore Algorithm, as the same algorithm was published by another researcher.

Before we get started, there are a couple of things that we must understand. Firstly, why does Dijkstra’s Algorithm fail for negative edge weights and second, the concept of Negative Cycles.

Why does Dijkstra fail?

Consider, the graph below,

Negative Edges in a Graph

Negative Edges in a Graph

The Dijkstra’s Algorithm is based on the principle that, if S → V1 → … → Vk is the shortest path from S → Vk then D(S, Vi) ≤ D(S, Vj). But in the above given graph, clearly that principle is violated. In the above graph, the shortest path from V1 to V3 is of length 3 units but the shortest path to V4 is of length 1 unit which means that V4 is actually closer to V1 than V3, which is contradicting Dijkstra’s principle.

Negative Cycles

A Negative Cycle is a path V1 → V 2 → V3 → … Vk → V1 where the total sum of the edge weights in the path is negative. Consider the graph below –

Negative Cycle in a Graph

Negative Cycle in a Graph

The path B → C → D is a negative cycle as the path’s total weight would be -2. So, the distance from A → B is 2, but if we circle the cycle once, we would get the distance as 0, if we circle once more, we would get -2. Like this we could keep on circling as much as we want to reduce the shortest distance. Hence the shortest distance to the vertex B, E becomes indeterminate.

So, we want Bellman Ford Algorithm to solve these two issues. We want it to compute the shortest path even for a graph with negative edges and negative cycles. The Bellman Ford will accurately compute the shortest path for a graph with negative edges but the algorithm fails for a graph with negative cycles. But, the algorithm can tell you if a negative cycle exists or not. If it exists the solution it puts up is incorrect, otherwise, the solution given by Bellman Ford Algorithm is perfect. This sounds fine because logically there will be no shortest paths for a graph with negative cycles.

Unlike the Dijkstra’s Algorithm where we had to use a priority queue, we will require no additional data structure to code Bellman Ford Algorithm. This makes writing the code much easier. And the algorithm is pretty straight-forward too. Take a look at the pseudo-code of the Bellman Ford Algorithm given below –

bellmanFord(G, s)
	for all edges in G(V)
		D(V) = INT_MAX
		parent[V] = -1

	D(s) = 0

	for i = 1 to |G(V)| - 1
		for each edge (u, v) in G(E)
			if edge can be Relaxed
				D(v) = D(u) + weight of edge (u, v)
				parent[v] = u

	for each edge in G(E)
		if edge can be Relaxed
			return false

	return true

You may not understand the pseudo-code at the first look, here’s a step-by-step representation of it –

  • Initialise the array which contains the shortest distances to infinity (a high integer value in the pseudo-code).
  • Initialise the parent array which contains the parent vertices in the shortest path to NULL (or -1 if it is an integer array).
  • Set the shortest distance of starting vertex to 0.
  • Explore all the edges, and see if you can relax them. If you can, relax the edge and proceed with the exploration.
  • Do the above operation |V| – 1 times.
  • After that, do another exploration on the graph checking all the edges if they can be relaxed. If they can be relaxed, you can a negative cycle in the graph. Hence, return false.
  • If the exploration gets finished successfully, the graph has no negative cycles and the data that you compute dis correct, so return true.

Now, what does exploring all the edges mean? If you are implementing the graph using an Adjacency List, it means to iterate over all the linked lists associated with all vertices. Now, what will be the sum of all the nodes in all Linked Lists in a given Adjacency List? Number of edges off course! So, we check all the edges from, edges of vertex 1 to vertex |V| in a linear manner. This whole operation takes O(|E|) time, which is repeated |V| – 1, so this part of the code takes O(|E||V|) time. Now, analyse the pseudo-code for a few minutes. Ask yourself how would you code this-ans-that. Now, when your mind is filled with the pseudo-code, look at the sketch below. The sketch below is sort of, “dry run” of the pseudo-code stated above –

Bellman Ford Algorithm Step-by-Step

Bellman Ford Algorithm Step-by-Step

The above sketch is self-explanatory. I hope you understand how the iterations go. In a way it looks like a very ordinary algorithm, without any greedy steps or partitions or so. The Bellman Ford Algorithm is pretty easy to code too. If you can work hard for an hour or two I’m sure you can code this algorithm. It does not require any priority queue or other tools. All you need to code Bellman Ford Algorithm is the pseudo-code. The pseudo-code is very important. Keep looking at the pseudo-code again-and-again whenever you get a doubt. I have put my code below for a reference, it is a C++ code –


If you have any doubts regarding the algorithm feel free to drop a comment. I’ll surely reply to them. I hope my post helped you in learning the Bellman Ford Algorithm. If it did, let me know by commenting! Keep practising… Happy Coding…! 🙂

Trie Tree Implementation

Hello people…! In this post we will talk about the Trie Tree Implementation. Trie Trees are are used to search for all occurrences of a word in a given text very quickly. To be precise, if the length of the word is “L“, the trie tree searches for all occurrences of this data structure in O(L) time, which is very very fast in comparison to many pattern matching algorithms.

But I must mention, this data structure is not exactly used for “pattern matching”, it is used to search for the occurrences of the word in the given text. How these both functionalities differ…? We’ll get to know that shortly. The Trie Tree has many applications. Your browser could be using a trie tree internally to search for words when you press Ctrl + F. So, let’s get started with this data structure…!

The Trie Tree is a very straight forward data structure. It is a simple tree where the nodes have an alphabet associated with them. And the way these nodes are arranged is the best part of the Trie Tree. To get an idea take a close look at the sketch below –

Structure of Trie Tree

Structure of Trie Tree

The arrows in the sketch above indicate how to traverse the trie tree in order to tell if a word exists or not. We travel through the root node, down the tree until we hit a leaf. Picking up a character at every edge, we construct our word. Now, you can easily tell why the time to search for a node in the text will be in the order of length of the word to be searched. It’s obvious…! One would have to go down till the leaf. In the worst case, one would have to travel the height of the tree which would be the length of the longest word in the whole text..! 😉

So, as we got a little idea about working with a Trie Tree, let us move on to the serious part. Let us talk about how this is implemented and then we will talk about three fundamental operations done on the Trie Tree –

  • Insert
  • Delete
  • Search

In C, the Trie Tree is implemented using structures. But the C implementation of this data structure gets a little dirty. So we will switch to C++ but we will keep it as C-ish as possible. As we keep discussing about the implementation, you will notice how many advantages we have when we use C++. This will be our structure of each node in the tree –

struct node
    struct node * parent;
    struct node * children[ALPHABETS];
    vector<int> occurrences;
  • parent – This points to the node which is the parent of the current node in the tree hierarchy. It may seem useless at the first, but we need this to travel up the tree when we want to delete any word. We’ll get to the deletion operation in a moment.
  • children – It points to the children nodes. It is made an array so that we can have O(1) accessing time. But why is the size 26…? Its simple. Consider a word, “th”, now, what could the third letter possibly be, if it had one…? One among the 26 english alphabets…! So that’s why the size is made 26. If you want to make a trie tree for another language, replace the number 26 with the number of letters in your language.
  • occurrences – This will store the starting indices of the word occurring in the given text. Now, why this is made a vector is that, vector is as good as a Linked List with random access. It is one of the most handy ready made data structures available in the C++ STL Library. If you are not familiar with vectors this is a good place to start.
    If this were C, we would have to give a fixed array size, and we would have a problem if the occurrences of a particular node are more. We could avoid this by putting a Linked List there. But we sacrifice random access and a whole lot of operations get time taking. Moreover, the code will get really really cumbersome to manage if you have a tree and a linked list.

Having got a picture of the implementation, let us look at how the operations are done in a Trie Tree.

Insert Operation

When we do an insert operation, there are a few cases –

  1. The word to be inserted does not exist.
  2. The word to be inserted already exists in the tree.
  3. The word to be inserted does not exists, but as the suffix of a word.

The first case is simple. One would have to traverse till the alphabets of the words have nodes in the trie tree or else create new nodes one-after-the-other. And at the end of the word, i.e., the node for the last alphabet, we will mark it as a leaf and push the starting index into the vector indicating the occurrence of the newly inserted word.

During this course of traversal, we will be cutting off the string of the word we have one-by-one as they are processed. This is done by putting using a vector of characters and popping off one character after the other. This is less code to handle and more efficient as we can use a vector as a queue. This is another advantage of using C++.

After having learnt what to do with the first case, you can guess what we would have to do in the second case. We simply have to add a new value to the occurrences vector at the node corresponding to the last alphabet of the word. We can also know the number of occurrences in constant time, we simply return the size of the vector. This is another advantage of using C++.

To understand the challenge in the third case, let’s take a simple example. What would you do with your trie tree if you wanted to insert the word “face” if the word “facebook” is already there in your tree…? This is the third case. The answer to this is the occurrence vector itself. We simply push the starting index of the word into the vector of that node which corresponds to the last alphabet of the word to be inserted, in the above example, this would be the node of “e”. So, what really tells if there’s a word ending with the alphabet corresponding to the current node is the size of the vector.

So I hope you understand how important our vector is. A lot depends on it…!

Delete Operation

The deletion of a word in the trie tree is similar to the insertion, we have a few cases –

  • Error 404 : Word not found…!
  • Word exists as a stand-alone word.
  • Word exists as a prefix of another word.

If the word is not there at all, we don’t have to do anything. We just have to make sure that we don’t mess up the existing data structure…!

The second case is a little tricky. We would have to delete the word bottom-up. That is, we will delete that part of the word which is not a part of any other word. For example, consider the sketch above. If we were to delete “this”, we would delete the letters ‘i’ and ‘s’ as, ‘h’ is a part of another word. This keeps us away from distorting the data structure. If the word were existing multiple number of times we will simply remove the occurrence from the vector of the concerned node.

In the third case too, we will simply delete the occurrence of the word from the vector. We needn’t write a lot of code as we can use the functions in algorithm header file. This is another advantage of using C++.

Note – When we delete the occurrence of a word, we are not concerned about the validity of the indices stored as occurrences of other words. What I mean to say is, suppose we have 10 words. If we delete the 3rd word, the 5th word or the 9th word is supposed to become the 4rth and the 8th word as far as the original text is concerned. But we will not consider this. The data stored in the trie tree is not meant to me deleted or inserted. The Trie Tree is meant for processing the given text not to manipulate the given text.

Search Operation

The search operation is simple and is as we discussed when we began our discussion about the Trie Tree. We go down the tree traversing the nodes and keep “picking up characters” as we go. And the occurrences vector tells us if a word exists that ends with the alphabet associated with the current node, and if so, it gives us the indices of occurrences and also the number of occurrences.

Besides these basic operations, there is another very interesting operation that is done with the Trie Tree –

  • Lexicographical Sort – If we want to print all the words processed into the trie tree lexicographically, all we have to do is do a Preorder Walk on the tree. This will automatically print all the words in the lexicographical order or the dictionary order. This is due to the very structure and arrangement of nodes in a Trie Tree. Now, I’d like to share another interesting thing about pre-order walk in trees… The Pre-order walk works exactly as a Depth First Search (DFS) in graphs, i.e., the sequence in which both the algorithms visit the nodes is the same. Think about this for a while and word out the two algorithms on an example (you could take mine in the sketch above), and you can see why it is so. You will also notice why the printed words would be lexicographically sorted.

Now, having learned a lot about the trie tree, try coding it in C++. If you are uneasy with C++, you can try it in C, but make sure you try at least 3 times. Trying is very important. I don’t know if you are new to reading my posts, but I insist a lot on trying in every post of mine…! If you have succeeded, you’re fabulous…! If not, check out my code below any try figuring out how just close you were…!!


The code is highly commented with explanation. It is well described, but if you have any doubts regarding the data structure or the code, feel free to comment them. I have used a few macros. The macro CASE indicates for which case the Trie Tree works. If we mention ‘A’ as the macro, the Trie Tree will work for upper case words only.

Other Implementations

The code is well tested against fairly large input. You can download the test case file here – Trie Tree Input (PDF). You can just clock your code for the insert operations. My code took 1.236 seconds to execute that loop which reads a word and inserts it into the Trie Tree. There are 5000 words in total. The last word in the input file is the word to be deleted.

If you think you can give a suggestion to make my code better, please do comment them too. I appreciate your suggestions. For those there who are struggling to get their code right, “Keep the effort going guys”…! Remember, you won’t learn anything if you keep writing Hello World program again and again. So, keep practising…! I hope my post has helped you in getting to know about the Trie Tree. If it did, let me know by commenting! Happy Coding…! 😀

Dijkstra’s Algorithm

Hello, people…! In this post, I will talk about one of the fastest single source shortest path algorithms, which is, the Dijkstra’s Algorithm. The Dijkstra’s Algorithm works on a weighted graph with non-negative edge weights and gives a Shortest Path Tree. It is a greedy algorithm, which sort of mimics the working of breadth first search and depth first search.

The Dijkstra’s Algorithm starts with a source vertex ‘s‘ and explores the whole graph. We will use the following elements to compute the shortest paths –

  • Priority Queue Q.
  • An array D, which keeps the record of the total distance from starting vertex s to all other vertices.

Just like the other graph search algorithms, Dijkstra’s Algorithm is best understood by listing out the algorithm in a step-by-step process –

  • The Initialisation –
  1. D[s], which is the shortest distance to s is set to 0. The distance from the source to itself is 0.
  2. For all the other vertices V, D[V] is set to infinity as we do not have a path yet to them, so we simply say that the distance to them is infinity.
  3. The Priority Queue Q, is constructed which is initially holds all the vertices of the graph. Each vertex V will have the priority D[V].
  • The Algorithm –
  1. Now, pick up the minimum priority (D[V]) element from Q (which removes it from Q). For the first time, this operation would obviously give s.
  2. For all the vertices, v, adjacent to s, i.e., check if the edge from s → v gives a shorter path. This is done by checking the following condition –

    if, D[s] + (weight of edge s → v) < D[v], we found a new shorter route, so update D[v]
    D[v] = D[s] + (weight of edge s → v)

  3. Now pick the next minimum priority element from Q, and repeat the process until there are no elements left in Q.

Let us understand this with the help of an example. Consider the graph below –

Dijkstra's Algorithm - Theory of ProgrammingFirstly, initialize your components, the shortest distances array D, the priority queue Q. The distance from the source to itself is zero. So, D[s] = 0, and the rest of the array is ∞. The set of vertices is inserted into the priority queue Q, with a priority D[v]. Now, we start our algorithm by extracting the minimum element from the priority queue.

Dijkstra's Algorithm - Theory of Programming

The minimum element in the priority queue will definitely be s (which is A here). Look at all the adjacent vertices of A. Vertices B, C, D are adjacent to A. We can go to B travelling the edge of weight 2, to C travelling an edge of weight 1, to D travelling an edge of weight 5. The values of D[B], D[C], D[D] are ∞ . We have found a new way of reaching them in 2, 1, 5 units respectively, which is less than ∞, hence a shorter path. This is what the if-condition mentioned above does. So, we update the values of D[B], D[C], D[D] and the priorities of B, C, D, in the priority queue. With this, we have finished processing the vertex A.

Dijkstra's Algorithm - Theory of Programming

Now, the process continues to its next iteration and we extract the minimum element from the priority queue. The minimum element would be vertex C which would be having a priority of 1. Now, look at all the adjacent vertices to C. There’s vertex D. From C, it would take 1 unit of distance to reach D. But to reach C in prior, you need 1 more unit of distance. So, if you go to D, via C, the total distance would be 2 units, which is less than the current value of shortest distance discovered to D, D[D] = 5. So, we reduce the value of D[D] to 2. This reduction is also called as “Relaxation“. With that, we’re done with vertex C.

Dijkstra's Algorithm - Theory of Programming

Now, the process continues to its next iteration and we extract the minimum element from the priority queue. Now, there are two minimum elements, B and D. You can go for anyone. We will go for vertex D. From D, you can go to E and F, with a total distance of 2 + 2 {D[D] + (weight of D → E)}, and 2 + 3. Which is less than ∞, so D[E] becomes 4 and D[F] becomes 5. We’re done with vertex D.

Dijkstra's Algorithm - Theory of Programming

Now, the process continues to its next iteration and we extract the minimum element from the priority queue. The minimum element in the priority queue is vertex B. From vertex B, you can reach F in 2 + 1 units of distance, which is less than the current value of D[F], 5. So, we relax D(F) to 3. From vertex B, you can reach vertex D in 2 + 2 units of distance, which is more than the current value of D(D), 2. This route is not considered as it is clearly proven to be a longer route. With that, we’re done with vertex B.

Dijkstra's Algorithm - Theory of ProgrammingNow, the process continues to its next iteration and we extract the minimum element from the priority queue. The minimum element in the priority queue is vertex E. From vertex E, you can reach vertex C in 4 + 4 units of distance, which is more than the current value of D(C), 1. This route is not considered as it is clearly proven to be a longer route. With that, we’re done with vertex E.

Dijkstra's Algorithm - Theory of ProgrammingNow, the process continues to its next iteration and we extract the minimum element from the priority queue. The minimum element in the priority queue is vertex F. You cannot go to any other vertex from vertex F, so, we’re done with vertex F.

Dijkstra's Algorithm - Theory of ProgrammingWith the removal of vertex F, our priority queue becomes empty. So, our algorithm is done…! You can simply return the array D to output the shortest paths.

Having got an idea about the overall working of the Dijkstra’s Algorithm, it’s time to look at the pseudo-code –

dijsktra(G, S)
    D(S) = 0
    Q = G(V)

    while (Q != NULL)
        u = extractMin(Q)
        for all V in adjacencyList[u]
            if (D(u) + weight of edge &amp;lt; D(V))
                D(V) = D(u) + weight of edge
                decreasePriority(Q, V)

In the pseudo-code, G is the input graph and S is the starting vertex. I hope you understand the pseudo-code. If you don’t, feel free to comment your doubts. Now, before we code Dijkstra’s Algorithm, we must first prepare a tool, which is the Priority Queue.

The Priority Queue

The Priority Queue is implemented by a number of data structures such as the Binary Heap, Binomial Heap, Fibonacci Heap, etc. The priority queue in my code is implemented by a simple Binary Min Heap. If you are not aware about the Binary Heap, you can refer to my post on Binary Heaps. Now the functionalities that we need from our priority queue are –

  • Build Heap – O(|V|) procedure to construct the heap data structure.
  • Extract Min – O(log |V|) procedure, where we return the top-most element from the Binary Heap and delete it. Finally, we make the necessary changes to the data structure.
  • Decrease Key – We decrease the priority of an element in the priority queue when we find a shorter path, as known as Relaxation.

If you know the working of the Binary Heap you can code the Priority Queue in about 1-2 hours. Alternatively, you can use C++ STL’s priority queue instead. But you don’t have a “decrease key” method there. So, if you want to use C++ STL’s priority queue, instead of removing elements, you can re-insert the same element with the lowered priority value.

Simple O(|V|2) Implementation

If you choose to implement priority queue simply by using an array, you can achieve the minimum operation in O(N) time. This will give your algorithm a total runtime complexity of O(|V|2). It is the simplest version of Dijkstra’s algorithm. This is the version you are supposed to use if you quickly want to code the Dijkstra’s algorithm for competitive programming, without having to use any fancy data structures. Take a look at the pseudocode again and try to code the algorithm using an array as the priority queue. You can use my code below as a reference –


Faster O(|E| log |V|) implementation

You can use a binary heap as a priority queue. But remember that to perform the decrease-key operation, you’ll need to know the index of the vertex inside the binary heap array. For that, you’ll need an additional array to store a vertex’s index. Each time any change is made in the binary heap array, corresponding changes must be made in the auxiliary array. Personally, I don’t like this version but I’ll put my code below so that you can use it as a reference.
If you want to do the same thing in C++, you can use a priority queue to reduce a lot. The tweak here is that because you cannot remove a certain vertex from a C++ STL priority queue, we can re-insert it with the new lower priority. This will increase the memory consumption but trust me, its worth it. I have put the codes for both versions below.

C using Binary HeapC++ using STL

The complexity of the above codes is actually O(|V| + |E|) ✗ O(log |V|), which finally makes it O(|E| log |V|). Dijkstra’s Algorithm can be improved by using a Fibonacci Heap as a Priority Queue, where the complexity reduces to O(|V| log |V| + |E|). Because the Fibonacci Heap takes constant time for Decrease Key operation. But the Fibonacci Heap is an incredibly advanced and difficult data structure to code. We’ll talk about that implementation later.

This is the Dijkstra’s Algorithm. If you don’t understand anything or if you have any doubts. Feel free to comment them. I really hope my post has helped you in understanding the Dijkstra’s Algorithm. If it did, let me know by commenting. I tried my best to keep it as simple as possible. Keep practising and… Happy Coding…! 🙂

Binary Heaps

Hello people…! In this post I will talk about Binary Heaps, or, more casually called as Heaps. Binary Heaps are used to implement Priority Queues which are associated with Scheduling. Binary Heaps is a starting step to learn many advanced heap related Data Structures that are used in many algorithms. So let’s understand the need of the situation by taking up a scenario.

Let’s say that the CPU has to execute ‘n’ programs that take their own pre-defined time to get completed. Then one strategy to schedule them is the Shortest Remaining Processing Time (SRPT) first method. As the name suggests, we look at all the available programs, then pick up that program which takes the minimum time to get executed. After completing that program, then the next minimum is supposed to be picked up. So if we observe what operations are we doing on the data we have, they would be –

  • Finding the minimum of given elements.
  • Deleting the minimum element (after a program gets executed).
  • Inserting a new element (if a new program enters the scheduling).

So, we must choose such a Data Structure which takes the minimum to these operations so that we can come up with an efficient solution to the scenario. Binary Heap is one such Data Structure that does these operations in pretty fast time. Now, what exactly is the Binary Heap…? What does it look like…? The Binary Heap is actually a Binary Tree has two very important properties –

  • Structural Property – The value of every node in the Binary Tree should be less than or equal to the values of its children nodes.
  • Heap Property – The Binary Tree should be fully-filled up to the 2nd last level and the last level must be “left-filled”.

To understand these two properties let’s take up an example –

Heap Property in Binary Heaps

Heap Property in Binary Heaps

So, as we can see, the last level is what is called “left-filled”, i.e., all the nodes are to the left and all the empty spaces to the right. The heap property is also depicted. The parent is smaller than both of its children. Now, having introduced this structure to you. Can you think where the smallest element of the heap would be….? At the top of course…! This is due to the very definition of the Heap Property. If the parent has the lower number, then the ancestor of all the nodes should surely have the least number. But this does not say that the leaves at the last level have the greatest number. Think about it for a while looking at the picture and you will understand it.

Now, the Binary Heap is although depicted as a tree, it is implemented as an array…! Just like the Fenwick Tree. The Binary Tree structure of the allows this. Look at the sketch below –

Storing a Binary Heap

Storing a Binary Heap

As, you can see, the nodes of the heap are put into an array in the left-to-right order at each level. Why is this so…? This is the way we can access all the nodes in a heap. How…? If we take any element arr[i] –

  • Parent of Element at i → arr[ i2 ]
  • Left Child of Element at → arr[(2 ✕ i)]
  • Right Child of Element at → arr[(2 ✕ i) + 1]

Looking at the picture, we can verify the above statements. For example, take the node 9 (arr[6]) which is at level 3. Its parent is arr[ 62 ] which is arr[3], i.e., node 6 at level 2, as it is in the structure. The left child of arr[6] is arr[(2 ✕ 6)], which is, arr[12], node 20 and like this even the right child can be verified. The way to find the parent is actually floor( i2 ), but as both ‘i’ and 2 are integers the result comes as it is desired. Now, we must discuss the five important operations in a binary heap –

  • Getting minimum element
  • Inserting a new element
  • Heapify
  • Deleting a new Element
  • Build Heap

Minimum Element – As you can clearly see, the minimum element would be the topmost element of the heap which is the first element of the array. So this operation takes O(1) time.

Inserting an Element – Recall that the last level of the heap must be left-filled. So, where ever you add a node, the resulting structure should be such that the last level will have an extra element. We do this, by first adding the new element to the end, then to ensure that the Structural Property is satisfied we keep swapping the element with its parent until the parent node to be swapped is less than the inserted element. So, we kind of insert the element at the bottom and we “move up” the heap. This process is illustrated in the sketch below.

Inserting an element into a Binary Heap

Inserting an element into a Binary Heap

This is how we insert a node into a heap. In the worst case, we would have to travel all the way up which costs us (log n) swaps, which is the height of the tree. So, the insertion operation has a complexity of O(log n).

Heapify – The heapify operation is as the name suggests, converting some “non-heap” thing into a heap. Well, what this non-heap thing is that it is a parent node, where the left and the right children are heaps, but the structure as a whole isn’t a heap. How is that possible…? This parent node we are talking about, its value could be greater than one or both of its children. In such a case, the smaller valued child is swapped with the parent and this is carried on down the tree until the heap property holds true for the whole structure. To understand this, observe the sketch below.

Heapify in Binary Min Heap

The process of heapify is clearly shown in the sketch, how we keep swapping the child nodes till the heap property is satisfied for the whole structure. This too takes O(log n) time in the worst case, when we would have to travel the whole till the leaves. Where is this process used anyway…? It is used to build a heap, which is our next discussion.

Deleting an Element – Deleting an element is a little tricky. Structurally, when an element is deleted, the last level will lose the right-most node. So what we do is, we swap the last element and the node to be deleted, then remove the last element. Based on the newly swapped element, we either bubble it up (as we did it in Insertion) or we bubble it down (which is simply to use heapify on the swapped element). Why would we get these two cases? Let us take some examples to understand this better. Study the sketch below –

Deletion in Binary Min Heap

So this is the case where we would have to call heapify on the swapped node. As we can see, this case arises when the swapped element is greater than both its children (greater than its parent). Now, the sketch below depicts the other case where we would have to bubble it upwards like we did it in insertion –


So this case arises when the swapped element is lesser than its parent. So, while deletion the conditions are –

if (swappedElement > parent) {
} else {
	bubble up swappedElement

Build Heap – Build heap is constructing the heap from a given set of randomly arranged nodes in a binary tree. The build heap process uses the heapify to construct the heap “down-to-up”. Now, if you remember the principle of the heapify process, we were able to do that only because the left-subtree and the right-subtree were heaps. Think of the second last level in a heap, the left and the right nodes are leaves, i.e., they have no children, so, they are heaps by themselves. Why is a leaf by itself a heap…? This is because the left child and the right child don’t exist, so we can claim that the parent node (the leaf), satisfies the heap property. And for a leaf, the structural property also holds good because the second-last level, (the level of the leaf node), is fully filled, and the last level (the level of the leaf’s children), is left-filled (actually there are no nodes to be left-filled). Well, most probably, you won’t understand this in the first reading, give it another couple of readings and observe the picture below.

Build Heap Step-by-Step

Build Heap Step-by-Step

I hope the process of the Build Heap is clear. At every level we use the heapify process to gradually build the heap “down-up”. So what is the asymptotic time complexity of Build Heap…? We use heapify for all the nodes except the nodes in the last level,

Nodes till the 2nd Last level = 12 ✕ (Total number of Nodes ⇒ N)
Complexity of the Heapify procedure = O(log N)
Complexity of Build Heap = O(N log N)

But we can argue that the Build Heap process will only take O(N). Now, look at the above sketch once again. If you feel the text is too small, open the image by right-clicking, its bigger than you think…! Now,

The (n + 1)4 nodes at level 3 took 1 swap ⇒ (n + 1)4 ✕ 1
The (n + 1)8 nodes at level 2 took 2 swaps ⇒ (n + 1)8 ✕ 2
The (n + 1)16 nodes at level 1 took 3 swaps ⇒ (n + 1)16 ✕ 3

As, you can see, we have a sort-of progression here. If we were to sum them up, taking (n + 1) common,

⇒ (N + 1) ✕ (14 + 28 + 316 + …)
⇒ (N + 1) ✕ Σ i2i
⇒ (N + 1) ✕ 2       { Σ i2i = 2 }
⇒ O(N)

Therefore, the Build Heap takes O(N) time. The principle behind this is that for most of the time Heapify works on smaller values of ‘N’. These are the most important operations of a Binary Heap. We have discussed the implementation of Binary Heaps using arrays and accessing the parent and child. Now, try to code the above-stated operations at least 3 times…! You have to try. There is no better way of knowing a Data Structure than trying to code it. After you have given your everything, if you succeed, you are brilliant…! If not, look at my code below and see for yourselves how close you were…!


One thing I would like to point out is that my heap is an array of structures. Just an integer would suffice, but I made it an array of structures so that it becomes a little generic. So, tomorrow, if you want to you this code as a Priority Queue, you needn’t change anything (or at most the names) in the code. Or, if you wanted to store more information about each node of the Binary Heap, you can simply add attributes to the struct.
Next, for my deletion operation, I swapped the last element and the element to be deleted and called the heapify procedure. This is because, the heapify procedure, checks if the Heap Property is violated and makes the necessary corrections. As the first element of the heap is the smallest, it can be accessed directly by heap[1]. Remember, the arrays in my code are 1-indexed, and all the loops and conditions go with it.
The code is pretty well tested. You can download the input file here.

There is one last topic to discuss before I conclude.

Max Heap – Whatever heaps we have been seeing till now are all Min Heaps. This is because the Heap Property is such that the minimum element is put on top. What if the Heap Property would just be the reverse of it…? Something like –

Heap Property – Every parent node must have a value greater than or equal to both of its child node.

In this case, by the Heap property, the maximum element would be put on the top and so will be for all the sub-trees or sub-heaps in the structure. This is a Max Heap. Actually, the very first heap that we used in our Build Heap sketch is a Max Heap. Have a close look at it and you will understand the difference. Max Heap differ from Min Heaps only by the Heap Property, due to which the other differences arise.

Whoo…! This has become a really long post…! Thanks a lot to stay with me till the end. I hope this helps you to get started with heaps. It may not be easy to code the Binary Heap, but that’s how you grow your skills in programming. You won’t learn anything if you sit for hours and write the Hello World program over-and-over again. You need to challenge yourself…! Feel free to comment your doubts..! Keep practicing…! Happy Coding…! 🙂

Depth First Search Algorithm

Hello people…! In this post I will talk about the other Graph Search Algorithm, the Depth First Search Algorithm. Depth First Search is different by nature from Breadth First Search. As the name suggests, “Depth”, we pick up a vertex S and see all the other vertices that can possibly reached by that vertex and we do that to all vertices in V. Depth First Search can be used to search over all the vertices, even for a disconnected graph. Breadth First Search can also do this, but DFS is more often used to do that. Depth First Search is used to solve puzzles! You can solve a given maze or even create your own maze by DFS. DFS is also used in Topological Sorting, which is the sorting of things according to a hierarchy. It is also used to tell if a cycle exists in a given graph. There are many other applications of DFS and you can do a whole lot of cool things with it. So, lets get started…!

The way the Depth First Search goes is really like solving a maze. When you see a maze in a newspaper or a magazine or anywhere else, the way you solve it is you take a path and go through it. If you find any junction or a crossroad, where you have a choice of paths to choose, you mark that junction, take up a path and traverse the whole route in your brain. If you see a dead end, you realize that this leads you now where so you come back to the junction you marked before and take up another path of the junction. See if that is also a dead end, else you continue to explore the, puzzle as this route contributes in reaching your destination. Well, at least that’s how I solve a maze. But… I hope you get the idea. You pick up a path and you explore it as much as possible. When you can’t travel any further and you haven’t yet reached your destination, you come back to the starting point and pick another path.

In code, you do this by recursion. Because of the very nature of recursion. Your recursion stack grows-grows and eventually becomes an empty stack. If you think for a while you can notice that the way of traversing which I told you above is logically covering only the vertices accessible from a given vertex S. Such traversal can be implemented using a single function that is recursive. But we want to explore the whole graph. So, we will use another function to do this for us. So, DFS is a two-functions thing. We will come back to the coding part later. Now, DFS too can be listed out in a step-by-step process –

  • For a given set of vertices V = {V1, V2,V3…}, start with V1, and explore all the vertices that can possibly be reached from V1.
  • Then go to the next vertex V2, if it hasn’t been visited, explore all the nodes reachable from V2, if not, go for V3.
  • Similarly, go on picking up all the vertices one-by-one and explore as much as possible if it wasn’t visited at all.

Now, how do you tell if a vertex wasn’t visited earlier…? If it has no parent vertex assigned. So what you have to do when you visit a node is –

  • Set the parent vertex of the current vertex to be the vertex from where you reached that vertex. We will use an array to assign parent vertices, which is initialised to -1, telling that the vertices were never visited.
  • When you are starting your exploration from a vertex, say, V1, we assign parent[V1] = 0, because it has no parent. If there is an edge from V1 to V2, we say, parent[V2] = V1.

Let’s look at an example and see how DFS really works –

Depth First Search Algorithm Step-by-Step

Depth First Search Algorithm Step-by-Step

The working of DFS is pretty clear from the picture. Notice how we would assign the parent vertices to each vertex. Once we have visited all the vertices from a given initial vertex V1, we backtrack to V1. What do we really mean by this “backtrack” here is that the recursion control will gradually come back to the function that started explopring from V1. We will understand this once we put DFS in code. And one more thing, whenever we got a choice of going to two vertices from one vertex, we preferred going to the vertex with the greater number. Why is this…? This is because we will be following Head Insertion in our Adjacency Lists to have O(1) Insertion operation. Now, assuming that we insert the vertices from Vertex 1 to Vertex 10, the greater number vertices will end up being in front of the Linked Lists. Take a moment to understand this. Even if you don’t understand, it’s ok…! You will get the hang of it later. Why I really did that was to explain you the concept of Edge Classification.

Edge Classification in Graphs

Edge Classification in Graphs

As you can see there are three types of edges, in fact, there are 4 actually. They are –

  • Tree Edge – These are the edges through which we have traversed all the vertices of the graph by DFS. More clearly, these are the edges that represent the parent-child relationship. That is, the tree edge Vertex 1 → Vertex 3 says that, Vertex 1 is the parent of Vertex 3. Just like the parent-child relationship in a tree. Why this is called a “tree” edge is that it happens so that these edges together form a “tree”, or rather a “forest”.
  • Forward Edge – This is an edge which points from one vertex which is higher in the hierarchy of parent-child relationship to a vertex which is a descendant. Observe that Vertex 2 is a descendant of Vertex 1, so the edge Vertex 1 → Vertex 3, is a forward edge.
  • Backward Edge – This is the opposite of forward edge. It points from a descendant Vertex to an ancestor Vertex. In the above diagram, the edge, Vertex 4 → Vertex 3 is a backward edge.
  • Cross Edge – Every other edge is a cross edge. We don’t have a cross edge in the above diagram, but, a cross edge can arise when there is a edge between siblings, two vertices that have the same parent.

Cycle Detection – Using DFS, we can detect if there are any cycles in the given graph. How…? This is very simple. If there exists at least one backward edge, then the given graph will have cycles. Well, telling how many cycles would be there for given number of backward edges is a challenge. But the point is if there is a backward edge, there is a cycle. This is by the very definition of the Backward Edge. It is an edge from a descendant to an ancestor. If we have a backward edge, then there will surely be another path of tree edges from the ancestor to descendant, forming a cycle. The picture below should make things clear.

Cycle Detection by Edge Classification

Cycle Detection by Edge Classification

But, how do we compute backward edges in code…? This is a little tricky. We could have a boolean array of size |V| which would hold the status of the vertex, whether it is in the recursion stack or not. If the vertex is in the recursion stack, it means that the vertex is indeed an ancestor. So, the edge will be a backward edge.

Now try to code the DFS, it is basically a recursion process. If you are good with recursion, I’m sure you can get this. Nonetheless, I have put my code below –


Just remember that, the parent array is used to indicate whether the vertex is visited or not, and it also indicates the path through which we came. On the other hand, the status array indicates, if a vertex is currently in the recursion stack. This is used to detect backward edges, as discussed. If a vertex has an edge which points to a vertex in the recursion stack, then it is a backward edge.

This is the Depth First Search Algorithm. It has a time complexity of O(|V| + |E|), just like the Breadth First Search. This is because, we visit every vertex once, or you could say, twice, and we cover all the edges that AdjacencyList[Vi] has, for all ViV which takes O(|E|) time, which is actually the for loop in our depth_first_search_explore() function. DFS is not very difficult, you just need to have experienced hands in recursion. You might end up getting stuck with some bug. But it is worth spending time with the bugs because they make you think in the perfect direction. So if you are not getting the code, you just have to try harder. But if you are having any doubts, feel free to comment them…! Keep practising… Happy Coding…! 🙂

Snakes and Ladders Game Code

Hello people…! In this post, we will discuss about the Snakes and Ladders Game Code, where we find the shortest path to win the Snakes and Ladders game by using the Breadth First Search (BFS) Algorithm. If you don’t know the algorithm, I suggest you read my post on Breadth First Search Algorithm.

Now, Graph Theory has many applications and I love working with things that have real-world applications, well, off course the other data structures too have their uses, but the speciality of Graph Theory is its applications have the closest association with our day-to-day activities. And to show you this and to set an example on how the BFS is actually put into action, I am taking up the example of the very popular game, Snakes and Ladders.

This game needs no introduction. We all must have played it in our childhood. I will explain you, how by using Graphs and BFS we can find the Shortest Path to win the game, and, we will state that path and the number of moves of dice it takes too. Have a good look at the Snakes and Ladder board below, we will be using it as an example throughout this post.

Snakes and Ladders

Snakes and Ladders

You can see we can reach the finish block by an number of ways, but how do you find out the best…? More importantly, how do you put it as code…? That’s what we are going to do. Now, think of how you can represent the game board in terms of a Graph, by Graph I mean in terms of Vertices and Edges.

Don’t be too hard on yourself… Just take the first 7 blocks and try working out on paper what would be the Edge and what would be the Vertices. If you ponder upon this, it is very easy to tell that the numbered blocks on the game board will be our Vertices, then, what will be the Edges…? This depends on how you can go from one block to another on rolling the dice. For now, forget about the ladders and the snakes, just draw the graph for a small portion of the board. It should look somewhat similar to what is in the picture below –

Dice Roll from Block 1

Dice Roll from Block 1

As you see we would have six ways to leave block 1, depending on the dice roll. Same would be the case for block 2, or, block 3, and so on. Now, there is a very important point to be noted here… Remember, this is Directed Graph…! Because, once you roll 5 and go to block 6, there’s no way to come back. So, the directions are important. Now, let us push up the complexity a bit by adding a ladder to our above sketch in block 6, think what would happen to the edges…?

Dice Roll for Block 1 - 5

Dice Roll for Block 1 – 5

If you roll 5 from block 1 you will jump directly to block 27. So is for block 2 when you roll out 4, or, block 3 when you roll out 3 and so on. Now, “logically” speaking, the block 6 does not exists in our graph…! Think about the statement for a while. Whenever you reach block, you are directly jumping to block 27, you don’t stay there. Now, if you were constructing an Adjacency List for this graph…. In the list of adjacent vertices for Vertex 1, would you have Vertex 6 in the list, or Vertex 27…? Vertex 27 of course…! Being at Vertex 6 means being at Vertex 27…!

That is why, our edge arrow did not end at Vertex 6… See it…? One more thing, in your Adjacency List, in the list of adjacent vertices for Vertex 6, what would you have…? Nothing…! Because you cannot come to a situation where you would have to stay on Vertex 6 and roll the dice. So the adjacent nodes list corresponding to Vertex 6 should be empty. These two things are very important, when you implement the Adjacency List for the Snake and Ladder board.

Same would be the case, if a snake was there at a block. Logically that block will not exist as a vertex in our adjacency list, and the corresponding edges must be removed. The only difference being that, the edge created due to a snake can lead you to a block of lower value.

Dice Roll for Block 25

Dice Roll for Block 25

Just to get a better idea of the scenario of a ladder and snake, I have depicted what the Adjacency List would look like for the above two examples shown in the pictures.

Adjustments to the Adjacency List

Adjustments to the Adjacency List

That pic should really clarify all the confusion about the graph. If you still don’t get it, feel free to comment your query…! Now, what do we do once we have the Adjacency List ready…? We just call the Breadth First Search method on that list…! Wait… Really…?! That’s it…? Yes…!

You see the hardest part here in solving the Snakes and Ladder by graphs is correctly determining what your Vertices and Edges are. Once you get that, all you have to do is the Breadth First Search in the resultant graph. Then you can get the shortest path from Vertex 1 to Vertex 100. Now, try getting the Adjacency Lit correct and simply call the BFS method. I’m sure you will succeed if you put in a little dedication. But for those who tried and failed. I have put my code below. Before you look at my code, there are a few logics I have used –

  • In the beginning of the program, I added all the edges as though the Game Board had no snakes or ladders at all (these number of edges is what I printed), then, I removed the respective edges concerning the snakes and ladders one-by-one.
  • When a Vertex ‘n’ has a ladder or a snake, we are supposed to replace the corresponding edges as I depicted in the pictures above, for that, I replaced the Vertex n‘s edge with the new value, in Vertices (n – 1), (n – 2), (n – 3), (n – 4), (n – 5), (n – 6). Because it is only in these vertices that you can find an edge of vertex n.
  • I put all this replacing stuff in a function replace() which takes the Linked List and searches for a value ‘oldVertex’ and replaces it with the value ‘newVertex’ when it finds it.
  • I used an extra element in my array to make them 1 – index based.
  • The number of moves to complete the shortest path would be the level of the last vertex, Vertex 100. Why…?! Think about it for a minute and you’ll get it…!
  • I have added a recursive function, printShortestPath() which recursively looks at the parent of each vertex until the start vertex is reached. It keeps printing vertices as the recursion stack keeps popping out, thus we get the path in a reverse order. Think about this for a while… Trust me… It is easy…! 😉

If you can solve this problem, then you can solve Snakes and Ladders: The Quickest Way Up problem of HackerRank. You can think of extending this to a 20 ✗ 20 Snakes and Ladder board or an nn board. You just have to add n2 edges to the graph. Solving puzzles by Graph Theory is real fun. I hope this post helped you. If you have any doubts, feel free to comment them. Keep practicing… and… Happy Coding…! 🙂

Breadth First Search Algorithm

Hello people…! In this post I will explain one of the most widely used Graph Search Algorithms, the Breadth First Search (BFS) Algorithm. Once you have learned this, you would have gained a new weapon in your arsenal, and you can start solving good number of Graph Theory related competitive programming questions.

What we do in a BFS is a simple step-by-step process, which is –

  1. Start from a vertex S. Let this vertex be at, what is called…. “Level 0”.
  2. Find all the other vertices that are immediately accessible from this starting vertex S, i.e., they are only a single edge away (the adjacent vertices).
  3. Mark these adjacent vertices to be at “Level 1”.
  4. There will be a challenge that you might be coming back to the same vertex due to a loop or a ring in the graph. If this happens your BFS will take time. So, you will go only to those vertices who do not have their “Level” set to some value.
  5. Mark which is the parent vertex of the current vertex you’re at, i.e., the vertex from which you accessed the current vertex. Do this for all the vertices at Level 1.
  6. Now, find all those vertices that are a single edge away from all the vertices which are at “Level 1”. These new set of vertices will be at “Level 2”.
  7. Repeat this process until you run out of graph.

This might sound heavy… Well atleast it would sound heavy to me if I heard it for the first time…! Many questions may pop-up in your mind, like, “How are we gonna do all that…?!”. Well, for now, focus on the concept, we will talk about the code later. And remember, we are talking about an Undirected Graph here. We will talk about Directed Graphs later. To understand the above stated steps, examine the picture below –

Breadth First Search Algorithm - Step-by-Step

Breadth First Search Algorithm – Step-by-Step

The sketch clearly shows you how we explore the vertices adjacent to a vertex and mark their levels. If you have noticed, whenever there were two ways of accessing the same vertex from multiple vertices of the same Level, i.e., in the diagram, Vertex 3 was accessible from Vertex 2 and Vertex 8, we preferred its parent to be Vertex 8, the larger number. Why is that so…? We will learn that in a short moment.

The concepts that I wish to emphasize from the above picture are, how BFS can tell you the shortest path from a given vertex (the start vertex) to all the other vertices and the number of edges or, the “length of the path”, would be the Level of that Vertex itself. This is a very important feature of the BFS, you will understand this more clearly when I explain it with the help of an example in a different post.

Now, having got some knowledge about the BFS, it is a good thing to do an exercise on this topic to really get the flow going. Try applying BFS on the Graph given. All you have to do is to implement the step-by-step process and get that final figure which I got above. And make sure you label the Levels and Parents for each vertex in the end.

Breadth First Search Practise Question

Breadth First Search Practise Question

Now, we come to the code part of the Breadth First Search, in C. We use the same Adjacency List that we used in our discussion of Graph Theory Basics. Coming back to our BFS discussion, the level of each vertex is stored in a separate array and so is the case for parent of each vertex. The three arrays are initialized to appropriate values. Now recall our step-by-step process that was stated earlier. Try to put that in terms of pseudocode and then proper code. Take a while to think how you would do that. If you could code it, you are marvelous…! 😀 … If not, don’t fret, I put my code below…!

CC++ STLC++ STL (using a Queue)Java

Try using the above example given as practice as the sample input to your code, so that you can easily compare the results. Now, the important point here is when we insert a vertex into the Adjacency List, we follow Head Insertion to get O(1) Insertion. What happens due to this is that, the Linked List will have numbers in the descending order.

So, when we are applying BFS for adjacent vertices of a given node, the Vertex with the higher number is met first. And then we explore starting by that higher numbered Vertex. This is why whenever we had a choice in approaching a vertex, we preferred approaching from the vertex which had the bigger number, why…? Due to Head Insertion…! If you had used Tail Insertion, this would be reversed.

Other Implementations of BFS

This is the overview of Breadth First Search Algorithm. It has a computational complexity of O(|V| + |E|), which is pretty fast. But why O(|V| + |E|)…? If you think about the algorithm a little bit, we cover all the |V| vertices level-by-level, checking all the |E| edges twice (for an Undirected Graph). Once the level is set for a subset of vertices we don’t visit them again. Put an itty-bitty thought into it and I’m sure you’ll get the idea…! 😉 … But, the code above runs slower than O(|V| + |E|)… To achieve the proper complexity we must use a Queue.

We can use BFS in the following scenarios –

  • Shortest Path or Quickest Path (if all edges have equal weight).
  • Finding the Length of Shortest Path between two Vertices.

With the knowledge of BFS you can start solving Graph Theory related problems. It really depends on your logic how you will apply the BFS to the given problem. And there is only one way to develop it… Practice…! So, keep practicing… Feel free to comment your doubts..! Happy Coding…! 🙂